Question

A quadratic function has x-intercepts at x=-6 and x=2. The point (-5,-14) is on the curve of the quadratic. What is the vertex of the quadratic?

Answer 1

To find the vertex of the quadratic function given the x-intercepts, we can use the fact that the vertex lies on the axis of symmetry of the parabola, which is halfway between the two x-intercepts.

Given the x-intercepts at \( x = -6 \) and \( x = 2 \), we can calculate the x-coordinate of the vertex as follows:

\[
\text{Vertex } x = \frac{-6 + 2}{2} = \frac{-4}{2} = -2
\]

Next, we need to find the y-coordinate of the vertex. We can express the quadratic function in vertex form or standard form. Since we have the x-intercepts, we can initially write the quadratic in factored form as:

\[
f(x) = a(x + 6)(x - 2)
\]

We need to determine the value of \( a \) using the additional point provided: (-5, -14).

Substituting \( x = -5 \) and \( f(x) = -14 \):

\[
-14 = a(-5 + 6)(-5 - 2)
\]

This simplifies to:

\[
-14 = a(1)(-7) \implies -14 = -7a \implies a = 2
\]

Now that we have \( a \), the quadratic function becomes:

\[
f(x) = 2(x + 6)(x - 2)
\]

Now we'll expand this to get it into standard form:

\[
f(x) = 2(x^2 + 4x - 12)
\]
\[
= 2x^2 + 8x - 24
\]

In standard form, the quadratic is \( f(x) = 2x^2 + 8x - 24 \).

To find the vertex, we can use the vertex formula for the x-coordinate:

\[
x = -\frac{b}{2a} = -\frac{8}{2 \cdot 2} = -\frac{8}{4} = -2
\]

Now substituting \( x = -2 \) back into the quadratic function to find the y-coordinate of the vertex:

\[
f(-2) = 2(-2)^2 + 8(-2) - 24
\]
\[
= 2(4) - 16 - 24
\]
\[
= 8 - 16 - 24 = -32
\]

Thus, the vertex of the quadratic function is:

\[
\boxed{(-2, -32)}
\]